Existence and Uniqueness Solutions for Some Strongly Quasilinear Parabolic Problems in Anisotropic Sobolev Spaces

Abstract

In this paper, we will study the strongly nonlinear parabolic problems of the type

$$\begin{aligned}\left\{ \begin{array}{ll} \displaystyle \frac{\partial b(u)}{\partial t} - \sum _{i=1}^{N} D^{i}(a_{i}(x,t,u,\nabla u))+ H(x,t,\nabla u) = f(x,t) \quad &{}\hbox { in }Q_{T}=\Omega \times (0, T),\\ u(x, t) = 0 &{}\hbox { on } S_{T}= \partial \Omega \times (0, T),\\ b(u(x, 0))= b(u_{0}(x)) &{}\hbox { in } \Omega , \end{array} \right. \end{aligned}$$

where $b(u_{0})$ belongs to $L^{1}(\Omega )$, and the behavior of $H(x,t,\xi )$ satisfies the growth conditions. Our work focuses on establishing the existence and uniqueness of a renormalized solution for this quasilinear parabolic equation. Furthermore, we conclude some regularity results.

Appendix

Lemma 6

Let $\displaystyle \psi _{h}(s)=1-\frac{\left| T_{2\,h}(s)- T_{h}(s)\right| }{h}$, then for any $h\ge 1$ we have

$$\begin{aligned} \displaystyle \int _{Q_{T}}\frac{\partial b_{n}(u_{n})}{\partial t}\psi _{h}(u_{n})(T_{k}(u_{n})-T_{k}(u))dx dt\ge \varepsilon _{4}(n,h). \end{aligned}$$

(121)

1.1 Proof of Lemma 6

Let $h\ge 1$, we define

$$\begin{aligned} \Psi _{h}(r)&=\displaystyle \int _{0}^{r}\psi _{h}(s)b_{n}'(s)ds\qquad \text{ and }\qquad \overline{\Psi }_{h,k}(r)=\int _{0}^{r}T_{k}(s)b_{n}'(s)\psi _{h}(s)ds. \end{aligned}$$

In view of (57), we have

$$\begin{aligned} \begin{array}{rl} &{}\displaystyle \lim _{\mu \rightarrow \infty }\liminf _{n\rightarrow \infty }\int _{Q_{T}}\frac{\partial \Psi _{h}(u_{n})}{\partial t}(T_{k}(u_{n})-T_{k}(u))dx dt \\ &{} \quad =\displaystyle \liminf _{n\rightarrow \infty }\int _{Q_{T}}\frac{\partial \Psi _{h}(u_{n})}{\partial t}T_{k}(u_{n})dx dt-\int _{Q_{T}}\frac{\partial \Psi _{k}(u)}{\partial t} T_{k}(u)dx dt\\ &{}\quad \displaystyle =\liminf _{n\rightarrow \infty }\int _{\Omega }\overline{\Psi }_{h,k}(u_{n}(T))-\overline{\Psi }_{h,k}(u_{0,n})dx-\int _{\Omega }\overline{\Psi }_{h,k}(u(T))-\overline{\Psi }_{h,k}(u_{0})dx\\ &{}\quad \displaystyle =\liminf _{n\rightarrow \infty }\int _{\Omega }\overline{\Psi }_{h,k}(u_{n}(T))dx-\int _{\Omega }\overline{\Psi }_{h,k}(u(T))dx. \end{array}\nonumber \\ \end{aligned}$$

(122)

On the other hand, it’s easy to see that $0\le \overline{\Psi }_{h,k}(r)\le B_{n,k}(r)$ and thanks to (32), then

$$\begin{aligned} \int _{\Omega }\overline{\Psi }_{h,k}(u_{n}(T))dx\le \int _{\Omega } B_{n,k}(u_{n}(T))dx\le C_{3} k. \end{aligned}$$

Since $\displaystyle \overline{\Psi }_{h,k}(u_{n}(T))\rightarrow \overline{\Psi }_{h,k}(u(T))$ a.e. in $\Omega$, thanks to the Fatou’s lemma we obtain

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _{\Omega }\overline{\Psi }_{h,k}(u(T))dx\le \liminf _{n\rightarrow \infty }\int _{\Omega }\overline{\Psi }_{h,k}(u_{n}(T))dx. \end{array} \end{aligned}$$

(123)

Hence

$$\begin{aligned} \begin{array}{rl} \displaystyle \liminf _{n\rightarrow \infty }\int _{Q_{T}}\frac{\partial \Psi _{h}(u_{n})}{\partial t}(T_{k}(u_{n})-T_{k}(u)))dx dt\ge 0. \end{array} \end{aligned}$$

(124)

Lemma 7

Let $u\in L^{\infty }(0,T,L^{2}(\Omega ))\cap L^{{\textbf {p}}}(0,T,W^{1,{\textbf {p}}}_{0}(\Omega )),$ then there exists a constant $C>0$ such that

$$\begin{aligned} \begin{array}{ll} \displaystyle \int _{Q_{T}}|u|^{{\overline{p}}\frac{N+2}{N}}dx dt \le C\Big (\sup _{t\in (0,T)}\int _{\Omega }|u|^{2}dx\Big )^{\frac{{\overline{p}}}{N}}\prod _{i=1}^{N}\Big (\int _{Q_{T}}|D^{i}u|^{p_{i}}dx dt\Big )^{\frac{{\overline{p}}}{Np_{i}}}. \end{array}\nonumber \\ \end{aligned}$$

(125)

1.1.1 Proof of Lemma 7

Let $t\in [0,T],$ by using Hölder’s and Sobolev inequalities together, we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _{\Omega }|u(x,t)|^{{\overline{p}}\frac{N+2}{N}}dx&{}\displaystyle \le \Big (\int _{\Omega }|u(x,t)|^{2}dx\Big )^{\frac{{\overline{p}}}{N}}\Big (\int _{\Omega }|u(x,t)|^{{\overline{p}}^{*}}dx\Big )^{\frac{N-{\overline{p}}}{N}}\\ &{}\displaystyle \le C\Big (\int _{\Omega }|u(x,t)||^{2}dx\Big )^{\frac{{\overline{p}}}{N}}\prod _{i=1}^{N}\Big (\int _{\Omega }|D^{i}u(x,t)|^{p_{i}}dx\Big )^{\frac{{\overline{p}}}{Np_{i}}}. \end{array}\nonumber \\ \end{aligned}$$

(126)

On the other hand, we set $\displaystyle f_{0}(t)=\int _{\Omega }|u(x,t)|^{2}dx,$ and $r_{0}=\frac{{\overline{p}}}{N},$ and for $i=1,\cdots ,N$

$$\begin{aligned} r_{i}=\frac{Np_{i}}{{\overline{p}}},\quad \hbox { and}\ \ f_{i}(t)=\Big (\int _{\Omega }|D^{i}u(x,t)|^{p_{i}}dx\Big )^{\frac{{\overline{p}}}{Np_{i}}}. \end{aligned}$$

We integrate in (126) on (0, T) and using Hölder’s inequality, we get

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _{0}^{T}\int _{\Omega }|u(x,t)|^{{\overline{p}}\frac{N+2}{N}}dx dt&{}\displaystyle \le C\int _{0}^{T}f_{0}(t)^{r_{0}}\prod _{i=1}^{N}f_{i}(t)dt\\ &{}\displaystyle \le C\sup _{t\in [0,T]}f_{0}(t)^{r_{0}}\int _{0}^{T}\prod _{i=1}^{N}f_{i}(t)dt\\ &{}\displaystyle \le C\sup _{t\in [0,T]}f_{0}(t)^{r_{0}}\prod _{i=1}^{N}\Big (\int _{0}^{T}f_{i}(t)^{r_{i}}dt\Big )^{\frac{1}{r_{i}}}. \end{array} \end{aligned}$$

(127)

Consequently (125) is hold.

Lemma 8

Assume that $Q_{T}=\Omega \times (0,T)$ with $\Omega$ is an open bounded subset of $I\!\!R^{N}$ and $p_{i}>1$ for $i=1,\cdots , N$. Let u be a measurable function satisfying

$$\begin{aligned} \displaystyle T_{k}(u)\in L^{\infty }(0,T,L^{2}(\Omega ))\cap L^{{\textbf {p}}}(0,T,W^{1,{\textbf {p}}}_{0}(\Omega )), \end{aligned}$$

for every $k>0$ and such that

$$\begin{aligned} \begin{array}{rl} \displaystyle \sup _{t\in (0,T)}\int _{\Omega }|T_{k}(u)|^{2}dx+\sum _{i=1}^{N}\int _{0}^{T}\int _{\Omega }|D^{i}T_{k}(u)|^{p_{i}}dx dt\le Mk, \end{array} \end{aligned}$$

(128)

where M is constant, then $|u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{N+{\overline{p}}}}\in L^{\frac{N+{\overline{p}}}{N},\infty }(Q_{T})$, and $|D^{i}u|^{\lambda _{i}}\in L^{\frac{N+2}{N+1},\infty }(Q_{T}),$ with

$$\begin{aligned} \displaystyle \lambda _{i}= \left\{ \begin{array}{lr} \frac{N(p_{i}-1)+p_{i}}{N+2}&{}\hbox { if}\quad p_{i}\le {\overline{p}},\\ \frac{N({\overline{p}}-1)+{\overline{p}}}{N+2}&{}\hbox { if}\quad p_{i}>{\overline{p}}, \end{array} \right. \qquad \hbox { and}\qquad \frac{1}{{\overline{p}}}=\frac{1}{N}\sum _{i=1}^{N}\frac{1}{p_{i}}. \end{aligned}$$

Moreover

$$\begin{aligned} \begin{array}{rl} \Vert |u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{N+{\overline{p}}}}\Vert _{ L^{\frac{N+{\overline{p}}}{N},\infty }(Q_{T})}\le CM, \end{array} \end{aligned}$$

(129)

and

$$\begin{aligned} \begin{array}{rl} \Vert |D^{i}u|^{\lambda _{i}}\Vert _{ L^{\frac{N+2}{N+1},\infty }(Q_{T})}\le CM+\hbox { meas}(Q_{T})^{\frac{N+1}{N+2}}. \end{array} \end{aligned}$$

(130)

1.1.2 Proof of Lemma 8

Firstly, we prove (129). Thanks to Lemma7 we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _{Q_{T}}|T_{k}(u)|^{{\overline{p}}\frac{N+2}{N}}dx dt &{}\le \displaystyle C\Bigg (\sup _{t\in (0,T)}\int _{\Omega }|T_{k}(u)|^{2}dx\Bigg )^{\frac{{\overline{p}}}{N}}\prod _{i=1}^{N}\Bigg (\int _{Q_{T}}|D^{i}T_{k}(u)|^{p_{i}}dx dt\Bigg )^{\frac{{\overline{p}}}{Np_{i}}}\\ &{} \displaystyle \le C(Mk)^{\frac{{\overline{p}}}{N}}(Mk)^{\frac{{\overline{p}}}{N} \sum _{i=1}^{N}\frac{1}{p_{i}}}\\ &{}\displaystyle \le C(Mk)^{\frac{{\overline{p}}}{N}+1}. \end{array}\nonumber \\ \end{aligned}$$

(131)

Let $k>0$, we have

$$\begin{aligned} \begin{array}{rl} \displaystyle k^{{\overline{p}}\frac{N+2}{N}}\hbox { meas }\{(x,t)\in Q_{T}: |u|>k\}&{}\displaystyle \le \int _{Q_{T}}|T_{k}(u)|^{{\overline{p}}\frac{N+2}{N}}dx dt\\ &{}\le \displaystyle C(Mk)^{\frac{{\overline{p}}}{N}+1}. \end{array} \end{aligned}$$

(132)

By taking $\displaystyle k=h^{\frac{N+{\overline{p}}}{N({\overline{p}}-1)+{\overline{p}}}}$, we get

$$\begin{aligned} \begin{array}{rl} \displaystyle h^{\frac{{\overline{p}}(N+2)(N+{\overline{p}})}{N(N({\overline{p}}-1)+{\overline{p}})}}\hbox { meas }\{(x,t)\in Q_{T}: |u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{{N+{\overline{p}}}}}>h\}&{}\displaystyle \le \int _{Q_{T}}|T_{k}(u)|^{{\overline{p}}\frac{N+2}{N}}dx dt\\ &{}\displaystyle \le C(Mh^{\frac{N+{\overline{p}}}{N({\overline{p}}-1)+{\overline{p}}}})^{\frac{{\overline{p}}}{N}+1}. \end{array}\nonumber \\ \end{aligned}$$

(133)

Hence

$$\begin{aligned} \begin{array}{rl} \displaystyle \hbox { meas }\{(x,t)\in Q_{T}: |u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{{N+{\overline{p}}}}}>h\}\le C(Mh^{-1 })^{\frac{{\overline{p}}}{N}+1}. \end{array} \end{aligned}$$

(134)

We obtain

$$\begin{aligned} \begin{array}{rl} \displaystyle h \hbox { meas }\{(x,t)\in Q_{T}: |u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{{N+{\overline{p}}}}}>h\}^{\frac{N}{N+{\overline{p}}}}\le CM\qquad \forall h>0. \end{array} \end{aligned}$$

(135)

Thus

$$\begin{aligned} \begin{array}{rl} \displaystyle \Vert |u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{{N+{\overline{p}}}}}\Vert _{L^{\frac{N+{\overline{p}}}{N},\infty }(Q_{T})}&{}\displaystyle =\sup _{h>0} h\hbox { meas}\{(x,t)\in Q_{T}: |u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{{N+{\overline{p}}}}}>h\}^{\frac{N}{N+{\overline{p}}}}\\ &{}\displaystyle \le CM. \end{array}\nonumber \\ \end{aligned}$$

(136)

Now, we prove that (130). First case $p_{i}<{\overline{p}}$, let $\gamma >1$ and using (128), we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \gamma ^{p_{i}} \hbox { meas }\{(x,t)\in Q_{T}:\ \ |D^{i}u|>\gamma \hbox { and } |u|\le k\}&{}\displaystyle \le \int _{\{|u|\le k\}}|D^{i}u|^{p_{i}}dx dt\\ &{}\displaystyle \le Mk. \end{array}\nonumber \\ \end{aligned}$$

(137)

According to (132) and taking $\displaystyle \gamma =\mu ^{\frac{N+2}{N(p_{i}-1)+p_{i}}}$, since $\displaystyle \gamma >1$ then $\mu >1$. Moreover we can see $\mu ^{\frac{p_{i}(N+2)}{N(p_{i}-1)+p_{i}}}\ge \mu ^{ \frac{{\overline{p}}(N+2)}{N({\overline{p}}-1)+{\overline{p}}}}$, therefore

$$\begin{aligned} \begin{array}{rl} \displaystyle meas\{(x,t)\in Q_{T}: |D^{i}u|^{\frac{N(p_{i}-1)+p_{i}}{N+2}}>\mu \}\le &{}\frac{Mk}{\mu ^{\frac{p_{i}(N+2)}{N(p_{i}-1)+p_{i}}}}+\frac{C(Mk)^{\frac{{\overline{p}}}{N}+1}}{k^{{\overline{p}}\frac{N+2}{N}}}\\ \displaystyle \le &{} \frac{Mk}{\mu ^{\frac{{\overline{p}}(N+2)}{N({\overline{p}}-1)+{\overline{p}}}}}+\frac{C(Mk)^{\frac{{\overline{p}}}{N}+1}}{k^{{\overline{p}}\frac{N+2}{N}}}. \end{array}\nonumber \\ \end{aligned}$$

(138)

Taking $\displaystyle k=M^{\frac{1}{N+1}}\mu ^{\frac{N+2}{N+1}\frac{N}{N({\overline{p}}-1)+{\overline{p}}}}$, we conclude that

$$\begin{aligned} \begin{array}{ll} \mu \> meas\{(x,t)\in Q_{T}: |D^{i}u|^{\frac{N(p_{i}-1)+p_{i}}{N+2}}>\mu \}^{\frac{N+1}{N+2}}\le CM \quad \text{ for } \text{ any }\ \mu >1. \end{array}\nonumber \\ \end{aligned}$$

(139)

It follows that

$$\begin{aligned} \begin{array}{rl} \Vert |D^{i}u|^{\lambda _{i}}\Vert _{L^{\frac{N+2}{N+1},\infty }(Q_{T})}&{}\displaystyle =\sup _{\mu> 0}\mu \hbox { meas }\{(x,t)\in Q_{T}: |D^{i}u|^{\frac{N(p_{i}-1)+p_{i}}{N+2}}>\mu \}^{\frac{N+1}{N+2}}\\ &{}\displaystyle \le \hbox { meas }(Q_{T})^{\frac{N+1}{N+2}}+\sup _{\mu> 1}\mu \hbox { meas }\{(x,t)\in Q_{T}: \\ &{}\quad |D^{i}u|^{\frac{N(p_{i}-1)+p_{i}}{N+2}}>\mu \}^{\frac{N+1}{N+2}}\\ &{}\displaystyle \le \hbox { meas }(Q_{T})^{\frac{N+1}{N+2}}+ CM. \end{array}\nonumber \\ \end{aligned}$$

(140)

For the second case of $p_{i}>{\overline{p}}$: Let $\gamma >1$, we have

$$\begin{aligned} \begin{array}{rl} \gamma ^{{\overline{p}}}\hbox { meas }\{(x,t)\in Q_{T}: |D^{i}u|>\gamma \hbox { and} |u|\le k\}&{}\displaystyle \le \int _{\{|u|\le k\}}|D^{i}u|^{{\overline{p}}}dx dt\\ {} &{}\displaystyle \le \int _{\{|u|\le k\}}|D^{i}u|^{p_{i}}dx dt\\ &{}\displaystyle \le Mk. \end{array}\nonumber \\ \end{aligned}$$

(141)

By taking $\gamma = \mu ^{\frac{N+2}{N({\overline{p}}-1)+{\overline{p}}}}$, we obtain

$$\begin{aligned} \begin{array}{rl} \displaystyle meas\{(x,t)\in Q_{T}: |D^{i}u|^{\frac{N({\overline{p}}-1)+{\overline{p}}}{N+2}}>\mu \} \le \frac{Mk}{\mu ^{\frac{{\overline{p}}(N+2)}{N({\overline{p}}-1)+{\overline{p}}}}}+\frac{C(Mk)^{\frac{{\overline{p}}}{N}+1}}{k^{{\overline{p}}\frac{N+2}{N}}}. \end{array}\qquad \end{aligned}$$

(142)

Similarly as in the first case, we deduce that

$$\begin{aligned} \begin{array}{rl} \Vert |D^{i}u|^{\lambda _{i}}\Vert _{L^{\frac{N+2}{N+1},\infty }(Q_{T})}\le \hbox { meas}(Q_{T})^{\frac{N+1}{N+2}}+ CM. \end{array} \end{aligned}$$

(143)

Lemma 9

Assume that $Q_{T}=\Omega \times (0,T)$ with $\Omega$ open subset of $I\!\!R^{N}$ of finite measure and $p_{i}>1$, for $i=1,\cdots , N$. Let u be a measurable function satisfying

$$\begin{aligned} \displaystyle T_{k}(u)\in L^{\infty }(0,T,L^{2}(\Omega ))\cap L^{{\textbf {p}}}(0,T,W_{0}^{1,{\textbf {p}}}(\Omega )), \end{aligned}$$

for every $k>0$ and $\frac{2(N+1)}{N+2}<\alpha \le {\underline{p}}$ such that

$$\begin{aligned} \begin{array}{rl} \displaystyle \sup _{t\in (0,T)}\int _{\Omega }|T_{k}(u)|^{2}dx\le Mk\quad \hbox { and }\quad \displaystyle \int _{0}^{T}\int _{\Omega }|\nabla T_{k}(u)|^{\alpha }dx dt\le C_{0}( Mk)^{\frac{\alpha }{2}}, \end{array}\nonumber \\ \end{aligned}$$

(144)

where M and $C_{0}$ is constant, then

$$\begin{aligned} \begin{array}{rl} \Vert u\Vert _{ L^{\frac{\alpha (N+2)}{2N},\infty }(Q_{T})}\le C M, \end{array} \end{aligned}$$

(145)

and

$$\begin{aligned} \begin{array}{rl} \Vert |\nabla u|\Vert _{ L^{\frac{\alpha (N+2)}{2(N+1)},\infty }(Q_{T})}\le C M, \end{array} \end{aligned}$$

(146)

where C is a constant depending only on N and $C_{0}$.

1.1.3 Proof of Lemma 9

since $\alpha <{\underline{p}}$, we have $\displaystyle L^{\textbf{p}}(0,T,W_{0}^{1,\textbf{p}}(\Omega )))\hookrightarrow L^{\alpha }(0,T,W_{0}^{1,\alpha }(\Omega )))$ is a continous embedding. By applying Gagliardo-Nirenberg and Höder inequality we get

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _{Q_{T}}|T_{k}(u)|^{\alpha \frac{N+2}{N}}dx dt &{}\le \displaystyle C\Bigg (\sup _{t\in (0,T)}\int _{\Omega }|T_{k}(u)|^{2}dx\Bigg )^{\frac{\alpha }{N}}\int _{Q_{T}}|\nabla T_{k}(u)|^{\alpha }dx dt\\ &{}\displaystyle \le C(Mk)^{\frac{\alpha (N+2)}{2N}} \end{array} \end{aligned}$$

(147)

It follows that: for every $k>0$ we have

$$\begin{aligned} \displaystyle k \hbox { meas }\{(x.t)\in Q_{T}: |u|>k\}^{\frac{2N}{\alpha (N+2)}}\le C M. \end{aligned}$$

(148)

Hence the estimation (145) hold true.

Now, we will prove (146). Using (147), we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \text{ meas }\{(x,t)\in Q_{T}: |\nabla u|>\gamma \}&{}{}\le \displaystyle \text{ meas }\{(x,t)\in Q_{T}: |\nabla u|>\gamma \text{ and }\> |u|\le k\}\\ {} &{}{}\displaystyle \qquad + \text{ meas }\{(x,t)\in Q_{T}: |\nabla u|>\gamma \> \text{ and }\> |u|>k\}\\ {} &{}{}\displaystyle \quad \le \frac{\int _{\{|u|\le k\}}|\nabla u|^{\alpha }dx dt}{\gamma ^{\alpha }}+\frac{C(Mk)^{\frac{\alpha (N+2)}{2N}}}{k^{\frac{\alpha (N+2)}{N}}}\\ {} &{}{}\displaystyle \quad \le \frac{C_{0}(Mk)^{\frac{\alpha }{2}}}{\gamma ^{\alpha }}+\frac{C(Mk)^{\frac{\alpha (N+2)}{2N}}}{k^{\frac{\alpha (N+2)}{N}}}. \end{array} \end{aligned}$$

(149)

By taking $k=\gamma ^{\frac{N}{N+1}}M^{\frac{1}{N+1}}$, we can conclude for any $\gamma >0$

$$\begin{aligned} \begin{array}{rl} \displaystyle \gamma \hbox { meas }\{(x,t)\in Q_{T}: |\nabla u|>\gamma \}^{\frac{2(N+1)}{\alpha (N+2)}}\le C M. \end{array} \end{aligned}$$

(150)

It follows that

$$\begin{aligned} \begin{array}{rl} \Vert |\nabla u|\Vert _{ L^{\frac{\alpha (N+2)}{2(N+1)},\infty }(Q_{T})}\le C M. \end{array} \end{aligned}$$

(151)